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This article is part of the series Proceedings of the International Congress in Honour of Professor Hari M. Srivastava.

Open Access Research

Particular solutions of a certain class of associated Cauchy-Euler fractional partial differential equations via fractional calculus

Shy-Der Lin*, Chia-Hung Lu and Shou-Mei Su

Author Affiliations

Department of Applied Mathematics, Chung Yuan Christian University, Chung-Li, 32023, Taiwan, R.O.C

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Boundary Value Problems 2013, 2013:126  doi:10.1186/1687-2770-2013-126


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2013/1/126


Received:30 January 2013
Accepted:1 May 2013
Published:16 May 2013

© 2013 Lin et al.; licensee Springer

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In recent years, various operators of fractional calculus (that is, calculus of integrals and derivatives of arbitrary real or complex orders) have been investigated and applied in many remarkably diverse fields of science and engineering. Many authors have demonstrated the usefulness of fractional calculus in the derivation of particular solutions of a number of linear ordinary and partial differential equations of the second and higher orders. The purpose of this paper is to present a certain class of the explicit particular solutions of the associated Cauchy-Euler fractional partial differential equation of arbitrary real or complex orders and their applications as follows:

where u = u ( x , t ) ; A, B, C, M, N, α and β are arbitrary constants.

MSC: 26A33, 33C10, 34A05.

Keywords:
fractional calculus; differential equation; partial differential equation; generalized Leibniz rule; analytic function; index law; linearity property; principle value; initial and boundary value

Dedication

Dedicated to Professor Hari M Srivastava.

1 Introduction, definitions and preliminaries

The subject of fractional calculus (that is, derivatives and integrals of any real or complex order) has gained importance and popularity during the past two decades or so, due mainly to its demonstrated applications in numerous seemingly diverse fields of science and engineering (cf.[1-15]). By applying the following definition of a fractional differential (that is, fractional derivative and fractional integral) of order ν R , many authors have obtained particular solutions of a number of families of homogeneous (as well as nonhomogeneous) linear fractional differ-integral equations.

In this paper, we present a direct way to obtain explicit solutions of such types of the associated Cauchy-Euler fractional partial differential equation with initial and boundary values. The results are a coincidence that the solutions are obtained by the methods applying the Laplace transform with the residue theorem. In this paper, we present some useful definitions and preliminaries for the paper as follows.

Definitions 1.1 (cf.[6-10])

If the function f ( z ) is analytic and has no branch point inside and on C , where

C : = { C , C + } (1.1)

C is an integral curve along the cut joining the points z and + i T ( z ) , C + is an integral curve along the cut joining the points z and + i T ( z ) ,

f ν ( z ) = C f ν ( z ) : = Γ ( ν + 1 ) 2 π i C f ( ζ ) d ζ ( ζ z ) ν + 1 ( ν R / Z ; Z : = { 1 , 2 , 3 , } ) (1.2)

and

f n ( z ) : = lim ν n { f ν ( z ) } ( n N : = { 1 , 2 , 3 , } ) , (1.3)

where ζ z ,

π arg ( ζ z ) π for  C , (1.4)

and

0 arg ( ζ z ) 2 π for  C + , (1.5)

then f ν ( z ) ( ν > 0 ) is said to be the fractional derivative of f ( z ) of order ν and f ν ( z ) ( ν < 0 ) is said to be the fractional integral of f ( z ) of order −ν, provided that

| f ν ( z ) | < ( ν R ) . (1.6)

First of all, we find it is worthwhile to recall here the following useful lemmas and properties associated with the fractional differ-integration which is defined above.

Lemma 1.1 (Linearity property)

If the functions f ( z ) and g ( z ) are single-valued and analytic in some domain Ω C , then

( k 1 f + k 2 g ) ν = k 1 f ν + k 2 g ν ( ν R , z Ω ) (1.7)

for any constants k 1 and k 2 .

Lemma 1.2 (Index law)

If the function f ( z ) is single-valued and analytic in some domain Ω C , then

( f μ ) ν = f μ + ν = ( f ν ) μ ( f μ 0 , f ν 0 , μ , ν R , z Ω ) . (1.8)

Lemma 1.3 (Generalized Leibniz rule)

If the functions f ( z ) and g ( z ) are single-valued and analytic in some domain Ω C , then

( f g ) ν = n = 0 ( ν n ) f ν n g n ( ν R , z Ω ) , (1.9)

where g n is the ordinary derivative of g ( z ) of ordern ( n N 0 : = N { 0 } ), it being tacitly assumed (for simplicity) that g ( z ) is the polynomial part (if any) of the product f ( z ) g ( z ) .

Lemma 1.4 (Cauchy’s residue theorem)

Let Ω be a simple connected domain, and letCbe a simple closed positively oriented contour that lies in Ω. Iffis analytic insideCand on C, expect at the point z 1 , z 2 , , z n that lie insideC, then

C f ( z ) d z = 2 π i k = 1 n Res [ f , z k ] .

(I) Iffhas a simple pole at z 0 , then

Res [ f , z 0 ] = lim z z 0 ( z z 0 ) f ( z ) ;

(II) Iffhas a pole of orderkat z 0 , then

Res [ f , z 0 ] = 1 ( k 1 ) ! lim z z 0 d k 1 d z k 1 ( z z 0 ) k f ( z ) .

Property 1.1 (cf.[6-10])

For a constanta,

( e a z ) ν = a ν e a z ( a 0 , ν R , z C ) . (1.10)

Proof The proofs between ‘ν is not an integer’ and ‘ν is an integer’ are not coincident, so we mention the proof as follows.

In case of | arg a | < π / 2 , we have

( e a z ) v = C ( e a z ) v = Γ ( v + 1 ) 2 π i C e a ζ ( ζ z ) v + 1 d ζ ( put  ζ z = η  and  a η = ξ , | arg η | π ) = a v e a z Γ ( v + 1 ) 2 π i e i ϕ ( 0 + ) ξ ( v + 1 ) e ξ d ξ ( ϕ = arg a ) = a v e a z Γ ( v + 1 ) 2 π i ( 0 + ) ξ ( v + 1 ) e ξ d ξ ( for  | ϕ | < π 2 ) = a v e a z

for | arg a | < π / 2 since ( 0 + ) ξ ( v + 1 ) e ξ d ξ = 2 π i Γ ( v + 1 ) .

In case of π / 2 < | arg a | π , we have

( e a z ) v = C + ( e a z ) v = Γ ( v + 1 ) 2 π i C + e a ζ ( ζ z ) v + 1 d ζ ( put  ζ z = η  and  a η = ξ , 0 arg η 2 π ) = a v e a z Γ ( v + 1 ) 2 π i e i ϕ ( 0 + ) ξ ( v + 1 ) e ξ d ξ ( ϕ = arg a ) = a v e a z Γ ( v + 1 ) 2 π i ( 0 + ) ξ ( v + 1 ) e ξ d ξ ( for  π 2 < | ϕ | π ) = a v e a z .

Therefore we have Property 1.1 for arbitrary a 0 . □

Property 1.2For a constanta,

( e a z ) ν = e i π ν a ν e a z ( a 0 , ν R , z C ) . (1.11)

Property 1.3For a constanta,

( z a ) ν = e i π ν Γ ( ν a ) Γ ( a ) z a ν ( ν R , z C , | Γ ( ν a ) Γ ( a ) | < ) . (1.12)

Property 1.4 (cf.[2,16,17])

The fractional derivative of a causal function f ( t ) is defined by

d α d t α f ( t ) : = { f ( n ) ( t ) if   α = n N , 1 Γ ( n α ) 0 t f ( n ) ( τ ) ( t τ ) α + 1 n d τ if   n 1 < α < n , (1.13)

where f ( n ) ( t ) denotes the ordinary derivative of ordernand Γ is the gamma function.

The Laplace transform of a function f ( t ) is denoted as

L { f ( t ) } ( s ) = 0 + e s t f ( t ) d t , (1.14)

wheresis the Laplace complex parameter. We recall from the fundamental formula (cf[16])

L { d α d t α f ( t ) } ( s ) = s α L { f ( t ) } ( s ) k = 1 n s α k f ( k 1 ) ( 0 ) , n 1 < α n , n N . (1.15)

2 Main results

Theorem 2.1The fractional partial differential equation

A x 2 2 u x 2 + B x u x + C u = M α u t α + N β u t β , (2.1)

with u = u ( x , t ) , n 1 < α , β n , n N andA (≠0), B, C, M, Nare constants, has its solutions of the form given by

(a)

u ( x , t ) = e λ t [ k 1 x A B + ( A B ) 2 4 A ( C M λ α N λ β ) 2 A + k 2 x A B ( A B ) 2 4 A ( C M λ α N λ β ) 2 A ] , (2.2)

when the discriminant ( A B ) 2 4 A ( C M λ α N λ β ) > 0 ;

(b)

u ( x , t ) = e λ t [ k 1 x m + k 2 x m ln x ] ,

when the discriminant ( A B ) 2 4 A ( C M λ α N λ β ) = 0 , and the roots m 1 , m 2 of Equation (2.5) are repeated; that is, m 1 = m 2 = m ;

(c)

u ( x , t ) = e λ t [ k 1 x a + b i + k 2 x a b i ] ,

when the discriminant ( A B ) 2 4 A ( C M λ α N λ β ) < 0 , and a + b i , a b i are the conjugate pair roots of Equation (2.5).

Proof Suppose that u ( x , t ) = x m e λ t . We have

u x = m x m 1 e λ t , 2 u x 2 = m ( m 1 ) x m 1 e λ t , α u t α = λ α x m e λ t , and β u t β = λ β x m e λ t . (2.3)

So that the given equation (2.1) becomes

A m ( m 1 ) x m e λ t + B m x m e λ t + C x m e λ t M λ α x m e λ t N λ β x m e λ t = 0 . (2.4)

Equation (2.4) leads to the auxiliary equation

A m ( m 1 ) + B m + C M λ α N λ β = 0 ( or  A m 2 ( A B ) m + ( C M λ α N λ β ) = 0 ) . (2.5)

That is,

m 1 = A B + ( A B ) 2 4 A ( C M λ α N λ β ) 2 A

and

m 2 = A B ( A B ) 2 4 A ( C M λ α N λ β ) 2 A

are the two roots of the auxiliary equation (2.5). Thus, u ( x , t ) = i = 1 2 k i x m i e λ t is a solution of the fractional partial differential equation (2.1) whenever m i ( i = 1 , 2 ) is a solution of the auxiliary equation (2.5).

There are three different cases to be considered, depending on whether the roots of this quadratic equation (2.5) are distinct real roots, equal real roots (repeated real roots), or complex roots (roots appear as a conjugate pair). The three cases are due to the discriminant of the coefficients ( A B ) 2 4 A ( C M λ α N λ β ) .

• Case I: Distinct real roots (when ( A B ) 2 4 A ( C M λ α N λ β ) > 0 ).

Let m 1 and m 2 denote the real roots of Equation (2.5) such that m 1 m 2 . Then the general solution of Equation (2.1) is

u ( x , t ) = e λ t ( k 1 x m 1 + k 2 x m 2 ) ,

where k i ( i = 1 , 2 ) are constants.

• Case II: Repeated real roots (when ( A B ) 2 4 A ( C M λ α N λ β ) = 0 ).

If the roots of Equation (2.5) are repeated, that is, m 1 = m 2 = m , then the general solution of Equation (2.1) is

u ( x , t ) = e λ t ( k 1 x m + k 2 x m ln x ) ,

where k i ( i = 1 , 2 ) are constants.

• Case III: Conjugate complex roots (when ( A B ) 2 4 A ( C M λ α N λ β ) < 0 ).

If the roots of Equation (2.5) are the conjugate pair m 1 = a + b i and m 2 = a b i , then a solution of Equation (2.1) is

u ( x , t ) = e λ t ( k 1 x a + b i + k 2 x a b i ) ,

where k i ( i = 1 , 2 ) are constants.

In general,

u ( x , t ) = e λ t [ k 1 x A B + ( A B ) 2 4 A ( C M λ α N λ β ) 2 A + k 2 x A B ( A B ) 2 4 A ( C M λ α N λ β ) 2 A ]

forms a fundamental solution, where k 1 , k 2 and λ are constants. □

Theorem 2.2The fractional partial differential equation

A 2 u x 2 + B u x + C u = M α u t α + N β u t β , (2.6)

with u = u ( x , t ) , n 1 < α , β n , n N andA (≠0), B, C, M, Nare constants, has its solutions of the form given by

(a′)

u ( x , t ) = e λ t [ k 1 e ( B + B 2 4 A ( C M λ α N λ β ) 2 A ) x + k 2 e ( B B 2 4 A ( C M λ α N λ β ) 2 A ) x ] , (2.7)

when the discriminant B 2 4 A ( C M λ α N λ β ) > 0 ;

(b′)

u ( x , t ) = e λ t [ k 1 e m x + k 2 x e m x ] ,

when the discriminant B 2 4 A ( C M λ α N λ β ) = 0 , and the roots m 1 , m 2 of Equation (2.10) are repeated; that is, m 1 = m 2 = m ;

(c′)

u ( x , t ) = e λ t [ k 1 x ( a + b i ) x + k 2 x ( a b i ) x ] ,

when the discriminant B 2 4 A ( C M λ α N λ β ) < 0 , and a + b i , a b i are the conjugate pair roots of Equation (2.10).

Proof The similarity between the forms of solutions of Equation (2.1) and solutions of a linear equation with constant coefficients of Equation (2.6) is not just a coincidence.

Suppose that u ( x , t ) = e m x e λ t . We have

u x = m e m x e λ t , 2 u x 2 = m 2 e m x e λ t , α u t α = λ α e m x e λ t , and β u t β = λ β e m x e λ t . (2.8)

So that the given equation (2.6) becomes

A m 2 e m x e λ t + B m e m x e λ t + C e m x e λ t M λ α e m x e λ t N λ β e m x e λ t = 0 . (2.9)

If m 1 and m 2 are the two roots of the auxiliary equation

A m 2 + B m + C M λ α N λ β = 0 , (2.10)

then

m = m 1 = B + B 2 4 A ( C M λ α N λ β ) 2 A

and

m = m 2 = B B 2 4 A ( C M λ α N λ β ) 2 A .

The analysis of three cases is similar to Theorem 2.1, we can obtain each solution of the forms as follows:

u ( x , t ) = e λ t ( k 1 e m 1 x + k 2 e m 2 x ) with m 1 m 2 , two distinct real roots,

u ( x , t ) = e λ t ( k 1 e m x + k 2 x e m x ) with m 1 = m 2 = m , repeated real roots, and

u ( x , t ) = e λ t ( k 1 e ( a + i b ) x + k 2 e ( a i b ) x ) with the conjugate complex roots.

 □

Remark The constant λ in Equations (2.2) and (2.7) can be solved directly by constant initial value and constant boundary values (or by the numerical methods).

Corollary 2.1The fractional partial differential equation

A x 2 2 u x 2 + B x u x + C = M α u t α , (2.11)

with u = u ( x , t ) , n 1 < α n , n N andA (≠0), B, C, Mare constants, has its solutions of the form given by

(a″)

u ( x , t ) = e λ t [ k 1 x A B + ( A B ) 2 4 A ( C M λ α ) 2 A + k 2 x A B ( A B ) 2 4 A ( C M λ α ) 2 A ] , (2.12)

when the discriminant ( A B ) 2 4 A ( C M λ α ) > 0 ;

(b″)

u ( x , t ) = e λ t [ k 1 x m + k 2 x m ln x ] ,

when the discriminant ( A B ) 2 4 A ( C M λ α ) = 0 , and the roots m 1 , m 2 of Equation (2.5) with N = 0 are repeated; that is, m 1 = m 2 = m ;

(c″)

u ( x , t ) = e λ t [ k 1 e a + b i + k 2 e a b i ] ,

when the discriminant ( A B ) 2 4 A ( C M λ α ) < 0 , and a + b i , a b i are the conjugate pair roots of Equation (2.5) with N = 0 .

Corollary 2.2The fractional partial differential equation

A 2 u x 2 + B u x + C u = M α u t α , (2.13)

with u = u ( x , t ) , n 1 < α n , n N andA (≠0), B, C, Mare constants, has its solutions of the form given by

( a )

u ( x , t ) = e λ t [ k 1 e ( B + B 2 4 A ( C M λ α ) 2 A ) x + k 2 e ( B B 2 4 A ( C M λ α ) 2 A ) x ] , (2.14)

when the discriminant B 2 4 A ( C M λ α ) > 0 ;

( b )

u ( x , t ) = e λ t [ k 1 e m x + k 2 x e m x ] ,

when the discriminant B 2 4 A ( C M λ α ) = 0 , and the roots m 1 , m 2 of Equation (2.10) with N = 0 are repeated; that is, m 1 = m 2 = m ;

( c )

u ( x , t ) = e λ t [ k 1 e ( a + b i ) x + k 2 e ( a b i ) x ] ,

when the discriminant B 2 4 A ( C M λ α ) < 0 , and a + b i , a b i are the conjugate pair roots of Equation (2.10) with N = 0 .

3 Examples

Example 3.1 If the two-dimensional harmonic equation 1 2 u = 0 is transformed to plane polar coordinates r and θ, defined by x = r cos θ , y = r sin θ , it takes the form

2 u r 2 + 1 r u r + 1 r 2 2 u θ 2 = 0 , (3.1)

then it has solutions of the form

u ( r , θ ) = ( k 1 r i λ + k 2 r i λ ) e λ θ ,

where k 1 , k 2 and λ are constants.

Solution Equation (3.1) is coincident to

r 2 2 u r 2 + r u r + 2 u θ 2 = 0 .

We have the solution

u ( r , θ ) = ( k 1 r i λ + k 2 r i λ ) e λ θ

by taking A = 1 , B = 1 , C = 0 , M = 1 , N = 0 and α = 2 in Theorem 2.1. □

Example 3.2 The fractional partial differential equation

2 u x 2 + u x = α u t α with  n 1 < α n , n N .

Solution Putting A = 1 , B = 1 , C = 0 and M = 1 in Corollary 2.2, we obtain the solution

u ( x , t ) = e λ t [ c 1 e ( 1 + 1 + 4 λ α 2 ) x + c 2 e ( 1 1 + 4 λ α 2 ) x ] ,

where the discriminant 1 + 4 λ α > 0 .

If 1 + 4 λ α = 0 ,

u ( x , t ) = e ( 1 2 x + 1 64 t ) ( c 1 + c 2 x ) .

The analysis of the case 1 + 4 λ α < 0 is similar to Theorem 2.2. □

Example 3.3 The fractional partial differential equation

x 2 2 u x 2 x u x = 2 1 2 u t 1 2 with  u ( 1 , t ) = u ( e , t ) = 0 .

Solution Putting A = 1 , B = 1 , C = 0 and M = 2 and α = 1 2 in Corollary 2.1, we obtain the solution

u ( x , t ) = x e λ t [ k 1 cos 4 × 2 λ 1 2 4 2 ln x + k 2 sin 4 × 2 λ 1 2 4 2 ln x ] = x e λ t [ k 1 cos 2 λ 1 2 1 ln x + k 2 sin 2 λ 1 2 1 ln x ] , u ( 1 , t ) = k 1 e λ t = 0 implies k 1 = 0 .

Then

u ( x , t ) = k 2 x e λ t sin 2 λ 1 2 1 ln x , u ( e , t ) = k 2 e λ t + 1 sin 2 λ 1 2 1 = 0 implies that 2 λ 1 2 1 = n π , n Z .

That is, λ = ( n 2 π 2 + 1 ) 2 4 .

Thus, u ( x , t ) = k 2 x e ( n 2 π 2 + 1 ) 2 4 t sin n π ln x , n Z .

If the discriminant 2 λ 1 2 1 = 0 , the solution is trivial. If the discriminant 2 λ 1 2 1 < 0 , then the solution is

u ( x , t ) = k 1 e ( n 2 + π 2 ) 2 t 4 ( x 1 + i n π x 1 i n π ) , n N .

 □

Example 3.4 The fractional partial differential equation

2 3 u t 2 3 = 1 4 2 u x 2 with  u ( 0 , t ) = e 2 t .

Solution Putting α = 2 3 , A = 1 4 , B = 0 , C = 0 and M = 1 in Corollary 2.2, the discriminant is λ 2 3 = ( λ 1 3 ) 2 0 , but λ = 0 leads to a contradiction, hence there are different real roots m 1 = 2 λ 1 3 and m 2 = 2 λ 1 3 , so that we have

u ( x , t ) = e λ t ( k 1 e 2 λ 1 3 x + k 2 e 2 λ 1 3 x ) .

By the boundary condition u ( 0 , t ) = e 2 t , we obtain λ = 2 and k 1 + k 2 = 1 . So,

u ( x , t ) = e 2 t e 2 2 3 x 2 k 1 e 2 t sinh ( 2 2 3 x ) = e 2 t e 2 2 3 x + 2 k 2 e 2 t sinh ( 2 2 3 x ) ,

and the particular solution is

u ( x , t ) = e 2 t e 2 2 3 x ( or  u ( x , t ) = e 2 t e 2 2 3 x ) .

If we apply the Laplace transform to ϕ ( t ) = u ( 0 , t ) = e 2 t , then L { ϕ ( t ) } ( s ) = 1 s + 2 and u ˜ ( x , s ) = 1 s + 2 e 2 x s 1 3 . Using the residue theorem,

u ( x , t ) = 1 2 π i C e s t 1 s + 2 e 2 x s 1 3 d s = 1 2 π i 2 π i lim s 2 ( s + 2 ) e 2 x s 1 3 e s t s + 2 = e 2 t e 2 2 3 x .

The solution obtained by the method of Laplace transform and the residue theorem is a coincidence, which is our result above. □

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

SDL carried out the molecular genetic studies, participated in the sequence alignment and drafted the manuscript. CHL carried out the immunoassays. SMS participated in the sequence alignment.

Acknowledgements

The authors are deeply appreciative of the comments and suggestions offered by the referees for improving the quality and rigor of this paper. The present investigation was supported, in part, by the National Science Council of the Republic of China under Grant NSC-101-2115-M-033-002.

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